Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $a = \dfrac{5z^2 + 5z}{z + 5} \times \dfrac{-2z^2 - 8z + 10}{z^3 - 6z^2 - 7z} $
Explanation: First factor out any common factors. $a = \dfrac{5z(z + 1)}{z + 5} \times \dfrac{-2(z^2 + 4z - 5)}{z(z^2 - 6z - 7)} $ Then factor the quadratic expressions. $a = \dfrac {5z(z + 1)} {z + 5} \times \dfrac {-2(z + 5)(z - 1)} {z(z + 1)(z - 7)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac {5z(z + 1) \times -2(z + 5)(z - 1) } {(z + 5) \times z(z + 1)(z - 7) } $ $a = \dfrac {-10z(z + 5)(z - 1)(z + 1)} {z(z + 1)(z - 7)(z + 5)} $ Notice that $(z + 1)$ and $(z + 5)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac {-10z(z + 5)(z - 1)\cancel{(z + 1)}} {z\cancel{(z + 1)}(z - 7)(z + 5)} $ We are dividing by $z + 1$ , so $z + 1 \neq 0$ Therefore, $z \neq -1$ $a = \dfrac {-10z\cancel{(z + 5)}(z - 1)\cancel{(z + 1)}} {z\cancel{(z + 1)}(z - 7)\cancel{(z + 5)}} $ We are dividing by $z + 5$ , so $z + 5 \neq 0$ Therefore, $z \neq -5$ $a = \dfrac {-10z(z - 1)} {z(z - 7)} $ $ a = \dfrac{-10(z - 1)}{z - 7}; z \neq -1; z \neq -5 $